Question

Evaluate: $\int \frac{2x+3}{\sqrt{x^2+4x+1}}dx$

Answer 1

$\int \frac{2x+3}{x^2+4x+1}dx=\int \frac{2x+3+1-1}{\sqrt{x^2+4x+1}}dx$

$=\int \frac{2x+4}{\sqrt{x^2+4x+1}}dx-\int \frac{1}{\sqrt{x^2+4x+1}}dx$

let $I_1=\int \frac{2x+4}{\sqrt{x^2+4x+1}}dx$ &

$I_2=\int \frac{1}{\sqrt{x^2+4x+1}}dx$

let $x^2+4x+1=t$

then $(2x+4)dx=dt$

so $I_1=\int \frac{dt}{\sqrt{t}}=2\sqrt{t}+C_1$

$I_1=2\sqrt{x^2+4x+1}+C_1$

$I_2=\int \frac{1}{\sqrt{x^2+4x+4-4+1}}dx$

$I_2=\int \frac{1}{\sqrt{(x+2{)}^2-(\sqrt{3}{)}^2}}dx$

$\because \int \frac{1}{\sqrt{x^2-a^2}=\ln |x+\sqrt{x^2-a^2}|}$

So $I_2=\ln |(x+2)+\sqrt{x^2+4x+1}|+C_2$

so, $I=I_1-I_2=2\sqrt{x^2+4x+1}-\ln |(x+2)+\sqrt{x^2+4x+1}|+C$